Basic because a pH greater than 7 is basic
Answer:
34g
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
H2S + 2AgNO3 —> 2HNO3 + Ag2S
Next, we shall determine the number of mole of H2S required to react with 2 moles of AgNO3.
This is illustrated below:
From the balanced equation above,
We can see that 1 mole of H2S is required to react completely with 2 moles of AgNO3.
Finally, we shall convert 1 mole of H2S to grams. This is shown below:
Number of mole H2S = 1 mole
Molar mass of H2S = (2x1) + 32 = 34g/mol
Mass = number of mole x molar Mass
Mass of H2S = 1 x 34
Mass of H2S = 34g
Therefore, 34g of H2S is needed to react with 2 moles of AgNO3.
Answer:
Pu-239
Explanation:
Beta decay moves the element which undergoes the decay one place to the right in the periodic table since to conserve charge and being beta radiations an electron we convert a neutron into a proton and an electron. In neutron capture we increase the atomic mas by one unit. We that in mind, lets solve the question:
U-238 + ₁⁰ n ⇒ U-239 ⇒ Np -239 + ₋₁⁰β ⇒ Pu-239 + ₋₁⁰β
The order of the steps, from first to last, is as follows based on the information in the question:
1. Salt is broken down into cations and anions.
2. Anion hydration.
3. Cation hydration.
4. Cations and anions that have been dissolved start to settle as a solid salt.
5. The rate of dissolution and recrystallization are equal.
Without hydration of the ions, the cation and anion cannot get separated. The three processes of ion dissociation, cation hydration, and anion hydration must all take place at once.
<h3>Why water is an effective solvent?</h3>
Water is a great solvent that can dissolve a wide variety of compounds due to its polarity and capacity to create hydrogen bonds.
Learn more about solvation here:
brainly.com/question/530845
#SPJ4
Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
